Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.1k views
in Basic Algebra by (47.1k points)
closed by

Solve for x2 – 7x3 + 8x2 + 8x – 8 = 0. Given 3 – √5 is a root.

1 Answer

+1 vote
by (49.3k points)
selected by
 
Best answer

When 3 – √5 is a root, 3 + √5 is the other root. 

S.o.r. = (3 – √5) + (3 + √5) = 6 

= (3 – √5) (3 + √5) = 32 - (√5)2

= 9 - 5 = 4

The equation is x2 – 6x + 4 = 0 

Now x4 – 7x3 + 8x2 + 8x – 8 = (x2 – 6x + 4) (x2 + px – 2) 

Equating co-eff of x 

12 + 4p = 8 

4p = 8 – 12 = -4 

So the other factor is x2 – x – 2

Now solving x2 – x – 2 = 0

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...